Maximize the probablity - tricky Gretel draws without replacement $n$ balls from the urn, which is $n$ white balls and $n$ black balls. White balls are numbered $1..n$, and black $n + 1, .. 2n$
We would like to Gretel to not draw balls, such that their number is $1$ or $2n$
That only thing we know is that she draw $b$ white balls and $n-b$ black.
Find such $b$ thatthe probability that Gretel didn't draw two special balls is greatest.
And my idea is following, but I don't know is it ok. Try to help me, please.
I will divide balls to four categories :
$Cat_1$ : white special ball (number one) was chosen
$Cat_2$ : $n-1$ white common balls were chosen
$Cat_3$ : $n-1$ black common balls were chosen
$Cat_4$ : black special ball (number $2n$) was chosen
$F_b$ - there were $b$ white balls chosen.
$F_{n-b} $ - there were $n-b$ black balls chosen.
$$P_r (F_b) = P_r(F_b|Cat_1)Pr(Cat_1) + P_r (F_b|Cat_2)Pr(Cat_2) $$
$$Pr(F_{n-b})=  P_r(F_{n-b}|Cat_3)Pr(Cat_3) + P_r (F_{n-b}|Cat_4)Pr(Cat_4) $$
So we search $Pr(Cat_4|F_{n-b}) + Pr(Cat_1|F_{b}) - Pr(Cat_4\cap Cat_1|  Pr(F_{n-b}\cap F_b) $
$$Pr(F_b | Cat_1) = \frac{{n-1\choose n-1}}{{2n-1\choose b-1}}$$
$$Pr(Cat_1) = \frac{1}{2n}$$
$$Pr(Cat_1|Fb) = \frac{Pr(F_b|Cat_1)Pr(Cat_1)}{Pr(F_b)} $$
Tell me, is it good way ?
 A: My interpretation of "the probability that Gretel didn't draw two special balls" is that Gretel can draw one or the other or none, just not both. 

I would not be casing on the categories. Since we are given the number of white and black balls drawn, I would case on that.
Given that we've drawn $b$ white balls out of $n$, the probability that the special white ball (ball $1$) is in those $b$ is $\frac{b}{n}$.
Given that we've drawn $n-b$ black balls out of $n$, the probability that the special black ball (ball $2n$) is in those $n-b$ is $\frac{n-b}{n}$.
We can also compute the probability that we haven't chosen the special ball.
Given that we've drawn $b$ white balls out of $n$, the number of ways not to draw the special white ball is $\binom{n-1}{b}$ out of $\binom{n}{b}$ ways to draw $b$ white balls.
Given that we've drawn $n-b$ black balls out of $n$, the number of ways not to draw the special black ball is $\binom{n-1}{n-b}$ out of $\binom{n}{n-b}$ ways to draw $n-b$ black balls.

Second Approach
Given that we've drawn $b$ white balls, the probability that a $1$ was not drawn is
$$
\frac{\binom{n-1}{b}}{\binom{n}{b}}=\frac{n-b}{n}
$$
and therefore, the probability that a $1$ was drawn is
$$
1-\frac{n-b}{n}=\color{#C00000}{\frac{b}{n}}
$$
Given that we've drawn $n-b$ black balls, the probability that a $2n$ was not drawn is
$$
\frac{\binom{n-1}{n-b}}{\binom{n}{b}}=\frac{b}{n}
$$
and therefore, the probability that a $2n$ was drawn is
$$
1-\frac{b}{n}=\color{#00A000}{\frac{n-b}{n}}
$$
Thus, the probability that we've drawn both a $1$ and a $2n$ is
$$
\color{#C00000}{\frac{b}{n}}\color{#00A000}{\frac{n-b}{n}}
$$
and therefore, the probability that we've avoided drawing both a $1$ and a $2n$ is
$$
1-\frac{b}{n}\frac{n-b}{n}
$$
Maximization with respect to $b$ is all that is left.
